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A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into the tank at a rate of 6 gal/min. the well-mixed solution is pumped out at the same rate. find the number a(t) of pounds of salt in the tank at time t.

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LammettHash
If [tex]A(t)[/tex] is the amount of salt in the tank at time [tex]t[/tex], then the rate at which the amount of salt in the tank changes is given by

[tex]\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}[/tex]
[tex]\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}[/tex]

Let's drop the units for now. We have

[tex]\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24[/tex]
[tex]e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}[/tex]
[tex]e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt[/tex]
[tex]e^{t/100}A(t)=2400e^{t/100}+C[/tex]
[tex]A(t)=2400+Ce^{-t/100}[/tex]

We're given that the water is pure at the start, so [tex]A(0)=0[/tex], giving

[tex]A(0)=0=2400+Ce^{-0/100}\implies C=-2400[/tex]

So the amount of salt in the tank (in lbs) at time [tex]t[/tex] is

[tex]A(t)=2400\left(1-e^{-t/100}\right)[/tex]

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